How to calculate pKa from Ka? - ka to pka, Conversion, Formulas, Equations

CH₃CH₂COOH is a weak acid that partially ionizes to yield H⁺ and CH₃CH₂COO^– ions in water.

As the K_a value for CH₃CH₂COOH is given in the question statement so we can find its pK_a by applying the equation as shown below.

∴ pK_a = -log₁₀K_a

∴ pK_a = -log₁₀(1.34 x 10^-5) = 4.87

Result: The pK_a value for propanoic acid (CH₃CH₂COOH) is 4.87.

HNO₃ is a strong acid that completely ionizes to give H⁺ and NO₃^– ions in an aqueous solution.

As the K_a value for HNO₃ is given in the question statement so we can find its pK_a by applying the equation as shown below.

∴ pK_a = -log₁₀K_a

∴ pK_a = -log₁₀(2.4 x 10¹) = -1.38.

Result: The pK_a value for nitric acid (HNO₃) is -1.38.

A negative pK_a signifies that HNO₃ is an extremely strong acid. The proton is only weakly held to the acid, which it readily liberates in an aqueous solution.

Butyric acid, also known as butanoic acid (CH₃CH₂CH₂COOH), is a weak acid that partially dissociates to release H⁺ and butanoate (CH₃CH₂CH₂COO^–) ions in an aqueous solution.

As the pH of the solution is given in the question statement, so we can find its [H⁺] at equilibrium using equation (vi).

∴ [H⁺]_equilibrium = 10^-pH ………. Equation (vi)

∴ [H⁺]_equilibrium = 10^-4.63 = 2.34 x 10^-5 M.

As per the balanced chemical equation shown above;

⇒ [H⁺]_equilibrium = [CH₃CH₂CH₂COO^–]_equilibrium

So [CH₃CH₂CH₂COO^–]_equilibrium = [H⁺]_equilibrium = 2.34 x 10^-5 M.

The initial butyric acid concentration is also given in the question, so

[CH₃CH₂CH₂COOH]_equilibrium = [CH₃CH₂CH₂COOH]_initial – [CH₃CH₂CH₂COO^–]_equilibrium

= 0.01 – (2.34 x 10^-5) = 9.98 x 10^-3 M.

Now that we know the acid dissociation constant (K_a) value, we can easily find pK_a by applying equation (i).

∴ pK_a = -log₁₀K_a…………. Equation (i)

∴ pK_a = -log₁₀(5.49 x 10^-8) = 7.26

Result: The pK_a value for the butyric acid solution is 7.26.

A comparatively high pK_a denotes a low acidic strength.

Example # 4: The K_a value for a weak acid is given to be 2.33 x 10^-11. Which of the following options provides the correct pK_a value for its acidic solution?

A) 11.23            B) 10.93           C) 10.63         D) 11.83        E) 9.93

Answer: Option C (pK_a = 10.63) is the correct answer.

Explanation: pK_a can be determined from the already given K_a value by applying equation (i).

pK_a = -log₁₀K_a …………. Equation (i)

pK_a = -log₁₀(2.33 x 10^-11) = 10.63

A weak acid (HA) partially dissociates to liberate H⁺ and A^– ions in water.

The concentration of hydrogen ions released at equilibrium is given in the question statement i.e., [H⁺] _equilibrium = 0.005 M.

∴ [H⁺] _equilibrium = [A^–]_equilibrium = 0.005 M

The original concentration of HA is also given in the question statement, so [HA]_equilibrium = [HA]_initial – [H⁺] _equilibrium = 0.025 – 0.005 = 0.02 M.

As a final step, we just need to substitute the above value into equation (i) to find the required pK_a value.

∴ pK_a = -log₁₀K_a …………. Equation (i)

∴ pK_a = -log₁₀(1.25 x 10^-3) = 2.90

Result: The pK_a value for the given acidic solution is 2.90.

What is K_a?

K_a stands for acid dissociation constant.

A weak acid (HA) partially dissociates to produce H⁺ and A^– ions in water. H⁺ ions combine with H₂O molecules to form hydronium (H₃O⁺) ions. A^– is known as the conjugate base of the acid. HA and A^– together are known as a conjugate acid-base pair.

The equilibrium constant for a reversible reaction is the ratio of the product of the concentration of products to the product of reactant concentrations.

The ionization equilibrium for the dissociation of HA in an aqueous solution can be represented as follows:

⇒ HA + H₂O ⇌ H₃O⁺ + A^–

The equilibrium constant (K_a) for the above reaction can be represented as equation (i)

⇒ Ka = \frac{[H_{3}O^{+}][A^{-}]}{[HA][H_{2}O]}………. Equation (i)

Where;

• [H₃O⁺] = concentration of hydronium ions formed in the aqueous solution
• [A^–] = concentration of conjugate base of the acid
• [HA] = acid concentration at equilibrium
• [H₂O] = concentration of water

As water concentration stays constant throughout the reaction, while [H₃O⁺] = [H⁺], i.e., the concentration of H⁺ ions released in the aqueous solution. So, equation (i) can be rearranged as equation (ii).

⇒ Ka = \frac{[H^{+}][A^{-}]}{[HA]}………. Equation (ii)

The greater the strength of an acid, it undergoes dissociation to a larger extent in the aqueous solution; thus, it possesses a higher K_a value.

Weak organic acids such as acetic acid and citric acid have K_a values below 1. However, strong mineral acids such as HCl and HNO₃ that completely dissociates to release a large number of H⁺ ions in an aqueous solution have K_a values above 1.

What is pK_a?

How is pK_a related to the strength of a Bronsted acid?

As per the Bronsted-Lowry acid-base theory, an acid is defined as a proton donor. The proton is loosely held to a strong acid which can readily liberate it in an aqueous solution.

However, weak acids strongly hold their protons, not ready to release in water.

Greater the pK_a value, the more strongly held the proton is by the Bronsted acid indicating a weak acidic strength and vice versa.

So pK_a is inversely related to the strength of a Bronsted acid.

How is K_a related to the strength of an acid?

Higher the K_a value, the greater the strength of an acid. So the acid dissociation constant is directly related to the strength of an acid.

Strong acids completely ionize in water to release H⁺ ions. Thus, they have high K_a values.

Weak acids only partially ionize in water to release a limited number of H⁺ ions, so they have low K_a values.

Also read: How to tell if an acid or base is strong or weak?

What is the relationship between pK_a and K_a?

What is the formula to calculate pK_a from K_a?

pK_a can be calculated by taking the negative logarithm of K_a as follows;

∴ pK_a = -log₁₀K_a.

How do you calculate K_a from pK_a?

K_a and pK_a are interconvertible chemical parameters. K_a can be calculated from pK_a by taking the antilog of pK_a as follows;

∴ K_a = 10^-pKa